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Let a = (1, 1, 0), b = (0, 1, 1), c = (1, 0, 1). The vector d = (2,…
Question
Let a = (1, 1, 0), b = (0, 1, 1), c = (1, 0, 1). The vector d = (2, 3, 5) is written uniquely as d = αa + βb + γc with real α, β, γ. Find α^2 + β^2 + γ^2.
✓ Verified answer: 13checked by our engine — not a guess
Step-by-step solution
Use the reciprocal-basis (Cramer) formulas: with V = [a b c], we have α = [d b c]/V, β = [a d c]/V, γ = [a b d]/V, equivalently α = d·(b×c)/V, etc.
V = det[a b c] = 2.
b × c = (1·1 - 1·0, 1·1 - 0·1, 0·0 - 1·1) = (1, 1, -1); d·(b×c) = 2 + 3 - 5 = 0 → α = 0.
c × a = (0·0 - 1·1, 1·1 - 1·0, 1·1 - 0·0) = (-1, 1, 1); d·(c×a) = -2 + 3 + 5 = 6 → β = 6/2 = 3.
a × b = (1·1 - 0·1, 0·0 - 1·1, 1·1 - 1·0) = (1, -1, 1); d·(a×b) = 2 - 3 + 5 = 4 → γ = 4/2 = 2.
(Check: 0·a + 3·b + 2·c = (2, 3, 5) = d.)
α^2 + β^2 + γ^2 = 0 + 9 + 4 = 13.
Final answer13
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